# Biophysics Problem 40

A \(70\;kg\) man and a \(0.5\; kg\) hairless dog go swimming. When they emerge \(0.5 \;kg\) of water adheres to the man's skin. How much adheres to the dog's? What percentage of the mass of each is water?

You will need \('L',\) the scaling factor which indicates how much greater the linear dimensions of the man are than those of the dog. You should first be able to find \(L^3\) from the information given in the question.

\(\frac{\text{mass of man}}{\text{mass of dog}} = \frac{\text{volume of man}}{\text{volume of dog}}= L^3 = \frac{70\;kg\;kg}{0.5kg\;kg}= 140\)

\(L\) will be the cube root of \(140,\) i.e. \(5.19.\)

Surface area scales as \(L^2,\) i.e., area is two-dimensional.

Using this information, what is the ratio of the surface area of the man to that of the dog?

Since surface area scales as \(L^2,\) then the ratio of the areas is \(L^2 = 27.\)

Now that we know the ratio of the areas \((27),\) we are ready to consider the ratio of the masses of water that would be required to *cover* these areas.

It is very important to realize that the thickness of the water layer will be the same on the man and the dog. This means that it will take \(27\) times as much water to cover the man as it takes to cover the dog.

Figure out the mass of the water on the dog.

This is a simple substitution:

\(\frac{\text{mass of man}}{\text{mass of dog}}=\frac{0.5\;kg}{\text{mass of dog}}= L^2= 27\)

From this, the mass of the dog is \(\mathrm{0.0185 \;kg.}\)

Therefore, the \(\mathrm{0.5\; kg}\) of the man is \(\mathrm{0.7\%}\) of his mass, and the \(\mathrm{0.0185 \;kg}\) of the dog is \(\mathrm{3.7\%}\) of its mass.